Talk:PlanetPhysics/Frobenius Method

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%%% Primary Title: Frobenius method
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\begin{document}

 Let us consider the linear homogeneous \htmladdnormallink{differential equation}{http://planetphysics.us/encyclopedia/DifferentialEquations.html} $$\sum_{\nu=0}^n k_\nu(x) y^{(n-\nu)}(x) = 0$$
of order $n$.\, If the coefficient \htmladdnormallink{functions}{http://planetphysics.us/encyclopedia/Bijective.html} $k_\nu(x)$ are continuous and the coefficient $k_0(x)$ of the highest order derivative does not vanish on a certain interval (resp. a domain in $\mathbb{C}$), then all solutions $y(x)$ are continuous on this interval (resp. domain).\, If all coefficients have the continuous derivatives up to a certain order, the same concerns the solutions.

If, instead, $k_0(x)$ vanishes in a point $x_0$, this point is in general a singular point.\, After dividing the differential equation by $k_0(x)$ and then getting the form
$$y^{(n)}(x)+\sum_{\nu=1}^n c_\nu(x)y^{(n-\nu)}(x) = 0,$$
some new coefficients $c_\nu(x)$ are discontinuous in the singular point.\, However, if the discontinuity is restricted so, that the products
$$(x-x_0)c_1(x),\quad (x-x_0)^2c_2(x),\quad \ldots,\quad (x-x_0)^nc_n(x)$$
are continuous, and even analytic in $x_0$, the point $x_0$ is a \htmladdnormallink{regular}{http://planetphysics.us/encyclopedia/CoIntersections.html} singular point of the differential equation.\\

We introduce the so-called\, {\em Frobenius method}\, for finding solution functions in a neighbourhood of the regular singular point $x_0$, confining us to the case of a second order differential equation.\, When we use the quotient forms
$$(x-x_0)c_1(x) := \frac{p(x)}{r(x)},\quad (x-x_0)^2c_2(x) := \frac{q(x)}{r(x)},$$
where $r(x)$, $p(x)$ and $q(x)$ are analytic in a neighbourhood of $x_0$ and\, $r(x) \neq 0$,\, our differential equation reads
\begin{align}
(x-x_0)^2r(x)y''(x)+(x-x_0)p(x)y'(x)+q(x)y(x) = 0.
\end{align}
Since a simple change\, $x\!-\!x_0\mapsto x$\, of variable brings to the case that the singular point is the origin, we may suppose such a starting situation.\, Thus we can study the equation
\begin{align}
x^2r(x)y''(x)+xp(x)y'(x)+q(x)y(x) = 0,
\end{align}
where the coefficients have the converging \htmladdnormallink{power}{http://planetphysics.us/encyclopedia/Power.html} series expansions
\begin{align}
r(x) = \sum_{n=0}^\infty r_nx^n,\quad p(x) = \sum_{n=0}^\infty p_nx^n,\quad q(x) = \sum_{n=0}^\infty q_nx^n
\end{align}
and
$$r_0 \neq 0.$$
In the Frobenius method one examines whether the equation (2) allows a series solution of the form
\begin{align}
y(x) = x^s\sum_{n=0}^\infty a_nx^n = a_0x^s+a_1x^{s+1}+a_2x^{s+2}+\ldots,
\end{align}
where $s$ is a constant and\, $a_0 \neq 0$.

Substituting (3) and (4) to the differential equation (2) converts the left hand side to
\begin{align*}
& [r_0s(s\!-\!1)\!+\!p_0s\!+\!q_0]a_0x^s+\\
& [[r_0(s\!+\!1)s\!+\!p_0(s\!+\!1)\!+\!q_0]a_1\!+\![r_1s(s\!-\!1)\!+\!p_1s\!+\!q_1]a_0]x^{s+1}+\\
& [[r_0(s\!+\!2)(s\!+\!1)\!+\!p_0(s\!+\!2)\!+\!q_0]a_2\!+\![r_1(s\!+\!1)s\!+\!p_1(s\!+\!1)\!+\!q_1]a_1\!+\![r_2s(s\!-\!1)\!+\!p_2s\!+\!q_2]a_0]x^{s+2}\!+\ldots
\end{align*}
Our equation seems clearer when using the notations\, $f_\nu(s) := r_\nu{s}(s\!-\!1)+p_\nu{s}+q_nu$:
\begin{align}
f_0(s)a_0x^s+[f_0(s\!+\!1)a_1+f_1(s)a_0]x^{s+1}+[f_0(s\!+\!2)a_2+f_1(s\!+\!1)a_1+f_2(s)a_0]x^{s+2}+\ldots = 0
\end{align}
Thus the condition of satisfying the differential equation by (4) is the infinite \htmladdnormallink{system}{http://planetphysics.us/encyclopedia/SimilarityAndAnalogousSystemsDynamicAdjointnessAndTopologicalEquivalence.html} of equations
\begin{align}
\begin{cases}
f_0(s)a_0 = 0\\
f_0(s\!+\!1)a_1+f_1(s)a_0 = 0\\
f_0(s\!+\!2)a_2+f_1(s\!+\!1)a_1+f_2(s)a_0 = 0\\
\qquad\cdots\qquad\cdots\qquad\cdots
\end{cases}
\end{align}
In the first place, since\, $a_0 \neq 0$,\, the {\em indicial equation}
\begin{align}
f_0(s) \equiv r_0s^2+(p_0-r_0)s+q_0 = 0
\end{align}
must be satisfied.\, Because\, $r_0 \neq 0$,\, this \htmladdnormallink{quadratic equation}{http://planetphysics.us/encyclopedia/QuadraticFormula.html} determines for $s$ two values, which in special case may coincide.

The first of the equations (6) leaves $a_0\,(\neq 0)$ arbitrary.\, The next linear equations in $a_n$ allow to solve successively the constants $a_1,\,a_2,\,\ldots$ provided that the first coefficients $f_0(s\!+\!1)$,\, $f_0(s\!+\!2),$\,$\ldots$ do not vanish; this is evidently the case when the roots of the indicial equation don't differ by an integer (e.g. when the roots are complex conjugates or when $s$ is the root having greater real part).\, In any case, one obtains at least for one of the roots of the indicial equation the definite values of the coefficients $a_n$ in the series (4).\, It is not hard to show that then this series converges in a neighbourhood of the origin.

For obtaining the complete solution of the differential equation (2) it suffices to have only one solution $y_1(x)$ of the form (4), because another solution $y_2(x)$, linearly independent on $y_1(x)$, is gotten via mere integrations; then it is possible in the cases\, $s_1\!-\!s_2 \in\mathbb{Z}$\, that $y_2(x)$ has no expansion of the form (4).

\begin{thebibliography}{9}
\bibitem{LP}{\sc Pentti Laasonen:} {\em Matemaattisia erikoisfunktioita}.\, Handout No. 261. Teknillisen Korkeakoulun Ylioppilaskunta; Otaniemi, Finland (1969).
\end{thebibliography} 

\end{document}
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